What's the probability that picking $3$ coins from a bag containing all coins from $1$ cent through $2$ euro will yield a total of $80$ cents?

I recently ran into this interview question and am wondering if my solution is correct. The setting is

We have a bag containing one coin of each type, i.e. we have one $$1$$ cent coin, one $$2$$ cent coin, one $$5$$ cent coin, one $$10$$ cent coin, one $$20$$ cent coin, one $$50$$ cent coin, one $$100$$ cent coin and one $$200$$ cent coin. We random select three coins from this bag, what is the probability that the sum of the values of the coins is at least $$80$$ cents? Start off with giving an initial guess for this probability, without computing anything, and explain your estimate.

My solution: For the estimation part, I figured there were $$\binom{7}{2}$$ configurations of three coins that contain the $$200$$ cent coin. Furthermore, there's also $$\binom{7}{2}$$ configurations of three coins that contain the $$100$$ cent coin. This already sums to $$21+21$$ configurations. However, this sum contains some configurations twice, so it is an overestimation. In total there are $$\binom{8}{3}=56$$ configurations of three coins. Thus the probability of taking three coins with a total value of atleast $$80$$ cents, is approximately $$\frac{42}{56}\approx\frac{2}{3}$$ (rough estimation, I know).

For the exact solution. I know that there are $$\binom{7}{2}$$ configurations of three coins that contain the $$200$$ cent coin, thus automatically satisfying that the sum is above $$80$$ cents. Then, to avoid duplicate configurations, there are $$\binom{6}{2}$$ configurations of coins containing the $$100$$ cent coin. Then there is one last configuration of $$50$$ cents, $$10$$ cents and $$20$$ cents, that also makes $$80$$ cents. Thus there are $$\binom{7}{2}+\binom{6}{2}+1=21+15+1=37$$ configurations. Thus the probability of taking atleast $$80$$ cents when randomly grabbing $$3$$ coins, is $$\frac{37}{56}$$.

Is this correct? Any help is appreciated.

• You really should list the values of all the coins explicitly. Not everybody lives in the Euro zone! Feb 16 '19 at 21:20
• For anyone curious, the euro coin denominations are 1 cent, 2 cents, 5 cents, 10 cents, 20 cents, 50 cents, 1 euro and 2 euro. (And 1 euro is 100 cents of course.) Feb 16 '19 at 21:23
• So it was sheer laziness? Feb 16 '19 at 21:37
• @TonyK I edited the post. Hope you can feel relaxed once more. Feb 16 '19 at 21:41
• @TonyK Fair enough, sorry. Feb 16 '19 at 22:00

To getting at least 80 cents:

we have 1€ and two from the rest: $${7\choose 2} =21$$

we don't have 1€, but we have 2€: $${6\choose 2} =15$$

we don't have 1€ and not 2€, then it is not possible if we have 10,20 and 50 cents

So we have 37 good possibilites among $${8\choose 3} = 56$$ so the probability is $$P= {37\over 56}$$

• That is incorrect for multiple reasons, but also because there are $8$ different coins in total. Feb 16 '19 at 21:03
• After the edit, that is also what I got as final answer. Thanks for the help! Feb 16 '19 at 21:10

More elegant (IMO) and easily doable with paper and pen:

there are $$~2~$$ ways to get at least $$~80c~$$

$$~1: 50+20+10 \to$$ probability is $$~3/8~$$ for first draw, $$~2/7~$$ for $$2$$nd draw, $$~1/6~$$ for third draw for total of: $$(3*2*1)/(8*7*6)$$

$$~2: 100 +~$$ anything or $$~ 200+$$anything

which is $$~1-(~$$Probability of not getting $$~100~$$ or $$~200)~$$

Probability of not getting $$~100~$$ or $$~200~$$ is $$~6/8~$$ for first draw, $$~5/8 ~$$ for $$2$$nd draw and $$~4/8~$$ for $$3$$rd draw

so total probability for option $$~2~$$ is $$1-(6*5*4)/(8*7*6)$$

So total probability to get at least $$~80c~$$ is

$$3*2*1/(8*7*6)+1-(6*5*4)/(8*7*6)$$ Simplify the fractions by dividing out the $$6:$$ $$1/(8*7)+1-5*4/(8*7)$$ $$=1+1/56-20/56$$ $$=(56+1-20)/56$$ $$=37/56$$